# Homework Help: Projectile problem

1. Sep 2, 2007

### munkachunka

1. The problem statement, all variables and given/known data

I have made an attempt at the following could someone please check my work.

a mortar shell fired with an angle of elevation of 52 deg has a range of 2.1km, neglecting air resistance, determine.

1, time of flight
2, initial velocity
3, shells range and height after 10 secs
4, the other range at which the height was the same as calculated in part 3 above

2. Relevant equations

x=u(cos(angle))t
h= u(sin, (angle))t-1/2gt^2

3. The attempt at a solution

a, rearanging x=u(cos52)t to equal x/cos52=ut
1hen x = 2.5km h=0

h=u(sin angle)t-1/2gt^2

substituting 0=(2.5knm/cos52)*(sin52)-1/2.9.8.t^2

transposing for t = 23.409

2, x=u(cos52)t therefore u=2100/(cos52)*23.409 = 145.714m/s

3= x=u(cos52)t = 145.712(cos52).10 = 897.093M

height = u(sin52)t-1/2gt^2
= 145.712*(sin52)*10-1/2*9.8*10^2

height = 658.226

4 assuming angle of firing = angle of descent and initial acceleration = rate of deceleration then the other range = 2.1km-897.093 = 1202.907

assuming the first 3 parts are correct then the part I am not too sure about is part 4, there are 12 marks allocated to this and so looks too easy.

your time is greatly appreciated

2. Sep 2, 2007

### Staff: Mentor

It looks good to me. (Just round off your answers to a reasonable number of significant figures.)

If you are unsure about part 4, solve it a second way and compare. (Use your equation for height as a function of time, solve for the time, plug into your equation for x to find the range.)

3. Sep 2, 2007

### learningphysics

Yeah, everything looks good... Just noticed this part:

the problem says 2.1km, you left a t out (probably nothing to do with your written work, but just typos)... but your calculations are right...

I noticed I was getting a few decimal places off from you in the answers... but other than that everything was the same.

Last edited: Sep 2, 2007
4. Sep 2, 2007

### munkachunka

thankyou for your time